5
$\begingroup$

I have been reading about the fundamental group and its connection to the first homology group. In fact, there is an isomorphism $$\pi_1^{ab}(X,x_0) \to H_1(X)$$ for every path-connected topological space $X$. So this provides a tool to compute the first homology group for a path-connected space when already knowing the fundamental group of that space just by dividing out the commutator subgroup.

I'm interested in a converse statement:

  1. Is it possible to compute the fundamental group $\pi_1(X,x_0)$ of a pointed space by knowing the homology group?
  2. In case yes, what are the assumptions to make?

Thank you for your help!

Tom

$\endgroup$
  • 1
    $\begingroup$ Only when it's abelian. If you know for some reason that your space has an abelian fundamental group (eg. it's an H-space), then you know that $\pi_1 X = H_1 X$. Otherwise... $\endgroup$ – Najib Idrissi Apr 24 '14 at 17:37
  • $\begingroup$ @NajibIdrissi Ok, of course I know that if $\pi_1$ is abelian and $X$ is path-connected, it is the same, which is somehow trivial. But are there any assumptions on $X$ or $H_1$ I can make to ensure the above isomorphism? Or another method of computing the fundamental group from the first homology group? $\endgroup$ – BIS HD Apr 24 '14 at 17:40
  • 1
    $\begingroup$ Are you asking under what conditions you have an isomorphism $\pi_1^{ab}(X,x) \to H_1(X)$? And no, you can't retrieve $\pi_1 X$ from $H_1 X$ otherwise. Just like you can retrieve $G$ from $G_{ab}$. Take the product of your space with $BG$ for any perfect group $G$, and you'll get a space with the same $H_1$ but a different $\pi_1$. $\endgroup$ – Najib Idrissi Apr 24 '14 at 17:42
  • 2
    $\begingroup$ In fact the non-commutativity of $\pi_1$ and the failure of the Hurewicz isomorphism can lead to very strange situations; for example there are spaces with trivial homology (meaning that $H_i X = 0$ for all $i > 0$) but that are not homotopically trivial. Such spaces (called acyclic spaces) have a perfect fundamental groups. $\endgroup$ – Najib Idrissi Apr 24 '14 at 17:44
  • 4
    $\begingroup$ Any condition would boil down to "the space has a commutative $\pi_1$". Sor for example "X is an H-space". As for why I think it's not possible, here's an example: the complement of any knot in $\mathbb{R}^3$ has an $H_1$ isomorphic to $\mathbb{Z}$, and yet the fundamental group of the complement is almost a complete invariant of the knot... $\endgroup$ – Najib Idrissi Apr 24 '14 at 17:56
7
$\begingroup$

There are some conditions that guarantee that the fundamental group of a space will be abelian. For example, if the fundamental group of an H-space is abelian. In these cases, the first homology group will be isomorphic to the fundamental group (if the space is path connected).

Otherwise, if you're only given the data of $H_1(X)$, you cannot compute $\pi_1(X)$ from that. The reason is simple: if you're given the abelianization of a group $G_{ab}$, the group $G$ could be pretty much anything. It could be the product of $G_{ab}$ with a perfect group, or some other extension of $G$...

To give you an example, a knot in $\mathbb{R}^3$ is almost determined by the fundamental group of its complement (as far as I remember, you also need to specify some orientation -- a knot theorist could correct me if I remember incorrectly). But it's also a theorem that the first homology group of this complement is always $\mathbb{Z}$! Even though knot complements are very well-behaved spaces, you still get a lot with the same first homology group.

In fact the noncommutativity of $\pi_1$ can lead to "strange" situations. For example, if the fundamental group is abelian, then trivial homology ($\tilde{H}_*(X) = 0$) implies trivial homotopy ($\pi_*(X) = 0$). But when the fundamental group is not abelian, then it's not true anymore, and there are in fact tons of so-called acyclic spaces whose homology vanish but who are not contractible. Their fundamental group will be perfect because of the Hurewicz isomorphism, but after that (almost) all bets are off. See for example Acyclic spaces by Dror Farjoun. So not even the whole homology of a space is enough to determine the fundamental group if you don't know that it's abelian.


Another example of possible condition is "$X$ is a co-H-space". The fundamental group of $X$ is then free, so the rank of the abelianization is enough to find $\pi_1$ up to isomorphism. I think you can even find the generators by considering lifts of generators of $H_1$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I know this answer was posted a long time ago (and it's a great answer!) but, regarding the knot theory comment, perhaps it is worth stating that a knot complement is determined by the map (induced by inclusion) $\pi_1(\partial K)\to \pi_1(K)$, where $\pi_1(\partial K)$ denotes the peripheral subgroup of $\pi_1(K)$ (the fundamental group of the torus boundary of the knot complement). I believe this is a result of Gordon-Luecke (for the unknot, it is still non-trivial - in this case, it is a corollary of the sphere theorem in 3-manifold topology). $\endgroup$ – Amitesh Datta Jan 3 '16 at 13:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.